SOLUTION: Given a+b=1, a^3+b^3=16. a). Expand (a+b)^3 b). Find the value of a^2 +b^2

Algebra ->  Equations -> SOLUTION: Given a+b=1, a^3+b^3=16. a). Expand (a+b)^3 b). Find the value of a^2 +b^2      Log On


   

Question 1109050: Given a+b=1, a^3+b^3=16.
a). Expand (a+b)^3
b). Find the value of a^2 +b^2
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
a). %28a%2Bb%29%5E3=a%5E3%2B3a%5E2b%2B3ab%5E2%2Bb%5E3

a%2Bb=1 --> system%28%28a%2Bb%29%5E3=1%2C%22and%22%2C+%28a%2Bb%29%5E2=1%29

Substituting into %28a%2Bb%29%5E3=a%5E3%2B3a%5E2b%2B3ab%5E2%2Bb%5E3 , we get
1=16%2B3a%5E2b%2B3ab%5E2
-15=3a%5E2b%2B3ab%5E2
-15=3ab%28a%2Bb%29
-15=3ab
-15%2F3=ab
ab=-5

a%5E2%2Bb%5E2%2B2ab=%28a%2Bb%29%5E2
Substituting %28a%2Bb%29%5E2=1%29 and ab=-5 , we get
a%5E2%2Bb%5E2%2B2%28-5%29=1
a%5E2%2Bb%5E2-10=1
a%5E2%2Bb%5E2=1%2B10
a%5E2%2Bb%5E2=11

NOTE: You could use system%28a%2Bb=1%2Cab=-5%29 to find
that a and b are %281+%2B-+sqrt%2821%29%29%2F2 ,
and then calculate a%5E2%2Bb%5E2 ,
but it is not fun.